hdu6370 并查集+dfs

Werewolf

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2363 Accepted Submission(s): 713

Problem Description

"The Werewolves" is a popular card game among young people.In the basic game, there are 2 different groups: the werewolves and the villagers.

Each player will debate a player they think is a werewolf or not.

Their words are like "Player x is a werewolf." or "Player x is a villager.".

What we know is :

1. Villager won't lie.

2. Werewolf may lie.

Of cause we only consider those situations which obey the two rules above.

It is guaranteed that input data exist at least one situation which obey the two rules above.

Now we can judge every player into 3 types :

1. A player which can only be villager among all situations,

2. A player which can only be werewolf among all situations.

3. A player which can be villager among some situations, while can be werewolf in others situations.

You just need to print out the number of type-1 players and the number of type-2 players.

No player will talk about himself.

Input

The first line of the input gives the number of test cases T.Then T test cases follow.

The first line of each test case contains an integer N,indicating the number of players.

Then follows N lines,i-th line contains an integer x and a string S,indicating the i-th players tell you,"Player x is a S."

limits:

Output

For each test case,print the number of type-1 players and the number of type-2 players in one line, separated by white space.

Sample Input

1 2 2 werewolf 1 werewolf

Sample Output

0 0

Source

2018 Multi-University Training Contest 6

思路：把狼边和人边分别记录，人边用并查集维护，最后判断狼边两端点是否在同一联通块，是则统计指向狼的人边的端点个数。

PS ：统计过程是由dfs和建反向边实现的。

代码：

 1 #include"bits/stdc++.h"
 2 
 3 #define db double
 4 #define ll long long
 5 #define vl vector<ll>
 6 #define ci(x) scanf("%d",&x)
 7 #define cd(x) scanf("%lf",&x)
 8 #define cl(x) scanf("%lld",&x)
 9 #define pi(x) printf("%d
",x)
10 #define pd(x) printf("%f
",x)
11 #define pl(x) printf("%lld
",x)
12 #define rep(i, n) for(int i=0;i<n;i++)
13 using namespace std;
14 const int N = 1e6 + 5;
15 const int mod = 1e9 + 7;
16 const int MOD = 998244353;
17 const db  PI = acos(-1.0);
18 const db  eps = 1e-10;
19 const ll  INF = 0x3fffffffffffffff;
20 int n,t;
21 int cnt,id;
22 struct P{int nx,to;}e[N];
23 struct PP{int fm,to;}g[N];
24 int head[N];
25 bool vis[N];
26 int fa[N];
27 char s[20];
28 int sum[N];
29 int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
30 void add(int u,int v)
31 {
32     e[id].to = u;
33     e[id].nx = head[v];
34     head[v] = id++;
35 }
36 void unio(int x,int y){
37     int xx,yy;
38     xx=find(x),yy=find(y);
39     if(xx!=yy) fa[xx]=yy;
40 }
41 
42 int dfs(int u)
43 {
44     vis[u]=1;
45     for(int i=head[u];~i;i=e[i].nx){
46         int v=e[i].to;
47         if(vis[v]) sum[u]+=sum[v];
48         else sum[u]+=dfs(v);
49     }
50     return sum[u];
51 }
52 int main(){
53     ci(t);
54     while(t--)
55     {
56         ci(n);
57         cnt=id=0;
58         memset(head,-1, sizeof(head));
59         memset(vis,0, sizeof(vis));
60         for(int i=0;i<n;i++) fa[i]=i,sum[i]=1;//初始化
61         for(int i=0;i<n;i++){
62             int u;
63             scanf("%d %s",&u,s);
64             u--;
65             if(s[0]=='w') g[cnt++]={i,u};
66             else{
67                 unio(i,u);
68                 add(i,u);//反向边
69             }
70         }
71         int ans=0;
72         for(int i=0;i<n;i++) if(!vis[i]) dfs(i);//统计每个点的子节点数
73         for(int i=0;i<cnt;i++){
74             if(find(g[i].fm)==find(g[i].to)) ans+=sum[g[i].to];
75         }
76         printf("0 %d
",ans);
77     }
78     return 0;
79 }